Quick question: Why do I need the $ in the following bits of code?
> main = withSocketsDo $ do > --do something with sockets > foo = fromMaybe 0 $ do > --do something in the maybe monad I don't see (after admittedly only a minute or so thinking about it) where any grammar ambiguities would be if 'do' had an implicit $ in front of it: > foo = fromMaybe 0 do > --do something in maybe Though now that I've written it down, that is hard for me to visually parse at a glance. I'm still curious though. --Jonathan _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
